Question

一枚公平的六面骰子掷两次。第一次掷出的数大于或等于第二次掷出的数的概率是多少？

A fair six-sided die is rolled twice. What is the probability that the first number that comes up is greater than or equal to the second number?

$\dfrac{1}{6}$

$\dfrac{5}{12}$

$\dfrac{1}{2}$

$\dfrac{7}{12}$

$\dfrac{5}{6}$

答案：D

知识点：骰子（概率）对称性

难度评级：1310

解答：

掷骰子两次时有 $3$ 类结果：第一次大于第二次、两次相等、第一次小于第二次。第一类和第三类由于对称而概率相同。

第二类发生的概率是 $\dfrac{1}{6}$ ，因为第一次可以是任意数，而第二次必须等于第一次。其余两类的总概率为 $1 - \dfrac{1}{6} = \dfrac{5}{6}$ 。因此每一类的概率为 $\dfrac{5}{6} \div 2 = \dfrac{5}{12}$ 。

所求概率是第一类加第二类，共为 $\dfrac{5}{12} + \dfrac{1}{6} = \dfrac{7}{12}$ 。

所以正确答案是 D。

There are $3$ possible outcomes when rolling a die twice: the first number is greater than the second, both numbers are equal, or the first number is less than the second number. The first and third outcomes have the same probability since they are symmetric.

The second outcome has a $\dfrac{1}{6}$ chance of happening, since the first number can be anything, and the second number must equal the first number. The other two outcomes have a combined probability of $1 - \dfrac{1}{6} = \dfrac{5}{6}.$ This means that each outcome has a $\dfrac{5}{6} \div 2 = \dfrac{5}{12}$ chance of happening.

The desired probability is the first outcome plus the second outcome, for a total probability of $\dfrac{5}{12} + \dfrac{1}{6} = \dfrac{7}{12}.$

Thus, D is the correct answer.

2011 AMC 8 第 18 题

答案：D

难度评级：1310

解答：

其他年份的第 18 题